Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can someone please explain to me how to prove this:

$\sin2\alpha = \frac{2\tan \alpha}{1+\tan^2 \alpha}$

Also: $\alpha \neq (2k + 1)\pi, k \in \mathbb{N}$

Thanks in advance.

share|improve this question

4 Answers 4

up vote 3 down vote accepted

$$\sin2\alpha=2\cos\alpha \sin\alpha=\dfrac{2\cos^2\alpha \sin\alpha}{\cos\alpha}=\dfrac{2\tan\alpha}{\sec^2\alpha}=\dfrac{2\tan\alpha}{1+\tan^2\alpha}$$

share|improve this answer

Use the fact that $\sin{2a}=2\sin{a}\cos{a}$. Now what happens if you multiply by 1?

$\mathbf{hint}: 1=\frac{\cos{a}}{\cos{a}}$...

share|improve this answer

Hint: Use the fact that $\sin 2\alpha = 2\sin \alpha \cos \alpha$ and $\sin^2 \alpha + \cos^2 \alpha = 1$. Use the last hint as the denominator of a fraction.

share|improve this answer

Right hand side= 2 $Tanx/(1+tan^2x)$ substitute Tanx=$sinx/cosx$ and 1+$tan^2x$= $sec^2x$

2$(sinx/cosx)$/$sec^2x$= 2$(sinx/cosx)$ * $cos^2x$ = 2sinx* cosx + sin2x (left hand side )

proven

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.